Question

If tan theta=3/4, find the value of 1-cos theta/1+cos theta

Answer 1

Given,

$tan( \theta) = \frac{3}{4}$

To find out,

$the \: value \: of \: \frac{(1 - \cos( \theta)) }{(1 + \cos( \theta)) }$

Solution:

$\tan( \theta) = \frac{opposite \: side \: to \: \theta}{adjacent \: side \: to \: \theta}$

$\tan( \theta) \: = \frac{3}{4}$

Therefore, opposite side:adjacent side =3:4.

In the figure,

For angle A,opposite side =BC= 3k

Adjacent side = AB=4k

(Note:where k is any positive number)

Now, we have in triangle ABC.

By pythagoras theorem.

${ac}^{2} = {ab}^{2} + {bc}^{2}$

${ac}^{2} = {3k}^{2} + {4k}^{2}$

${ac}^{2} = 25 {k}^{2}$

$ac \: = \sqrt{25 {k}^{2} }$

$ac \: = 5k$

Therefore the hypotenuse is 5k.

Now,

$\cos( \theta) = \frac{adjacent \: side \: to \: \theta}{hypotenuse}$

$\cos( \theta) = \frac{3k}{5k} = \frac{3}{5}$

$\frac{(1 - \cos( \theta)) }{(1 + \cos( \theta)) } = \frac{(1 - \frac{3}{5}) }{(1 + \frac{3}{5}) }$

$\frac{ \frac{5 - 3}{5} }{ \frac{5 + 3}{5} }$

$\frac{ \frac{2}{5} }{ \frac{8}{5} }$

$\frac{2}{5} \times \frac{5}{8}$

$\frac{1}{4}$

Therefore the value is 1/4.